Analysis Problem of the Day 31

Today’s problem appeared as Problem 9 on the UCLA Fall 2011 Analysis Qual:

Problem 9: Let $f$ be a function holomorphic on the punctured unit disk $\mathbb{D}^*:=\mathbb{D} \setminus \{0\},$ and suppose that $f$ is also square integrable, i.e. $\int_{\mathbb{D}^*} |f(z)|^2 d\mu(z)<\infty.$ Show that $f$ extends to a holomorphic function on the entire unit disk $\mathbb{D}.$

Solution: Since $f$ is holomorphic on the annulus $0<r<1,$ let’s expand $f$ in a Laurent series around $z=0,$ i.e $f(z) = \sum_{n=-\infty}^\infty a_n z^n,$ which converges to $f$ normally on $\mathbb{D}^*.$ Additionally, since we are working with the $L^2$ norm of the function, let’s rewrite it in polar coordinates, i.e. $z=re^{i\theta}$ and $f(z) = \sum_{n=-\infty}^\infty a_n r^n e^{in\theta}.$ Then, the condition that $f$ is square integrable can be rewritten as

$\int_0^{2\pi} \int_0^1 \left(\sum_{n=-\infty}^\infty a_n r^n e^{in\theta} \right) \left(\sum_{m=-\infty}^\infty \overline{a_m} r^m e^{-im\theta} \right) r dr d\theta = 2\pi \sum_{n=-\infty}^\infty |a_n|^2 \int_0^1 r^{2n+1} dr<\infty,$

where we can interchange the integral and sum freely by Fubini and where we use the fact that $\int_0^{2\pi} e^{in\theta} = 2\pi \delta_{n,0}.$ Notice, in particular, that $r^{2n+1}$ is integrable near the origin if and only if $2n+1>-1,$ i.e. $n>-1.$ Thus, the Laurent series for $f$ has no negative terms, meaning $f$ has a removable singularity at the origin and thus may be extended to a holomorphic function on the entire unit disk.

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